question stringlengths 37 4.33k | thinking stringlengths 2.04k 2.53k | answer stringlengths 1 261 | id int64 56 93.7k | content stringlengths 2.04k 2.77k | id_ stringlengths 3 6 |
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Evaluate $\sec'' \frac{\pi}4 +\sec'' \frac{3\pi}4+\sec'' \frac{5\pi}4+\sec'' \frac{7\pi}4$. (Here $\sec''$ means the second derivative of $\sec$). | <think>1. First, we need to find the second derivative of the secant function, $\sec x$. We start by finding the first derivative:
\[
\frac{d}{dx} \sec x = \sec x \tan x
\]
2. Next, we find the second derivative by differentiating the first derivative:
\[
\frac{d^2}{dx^2} \sec x = \frac{d}{dx} (\sec x \... | 0 | 49,720 | <think>1. First, we need to find the second derivative of the secant function, $\sec x$. We start by finding the first derivative:
\[
\frac{d}{dx} \sec x = \sec x \tan x
\]
2. Next, we find the second derivative by differentiating the first derivative:
\[
\frac{d^2}{dx^2} \sec x = \frac{d}{dx} (\sec x \... | 1_0 |
Determine all triples of natural numbers $(a,b, c)$ with $b> 1$ such that $2^c + 2^{2016} = a^b$. | <think>We are given the equation \(2^c + 2^{2016} = a^b\) and need to determine all triples of natural numbers \((a, b, c)\) with \(b > 1\).
### Lemma:
The equation \(2^b + 1 = x^a\) with \(x > 0\) and \(a, b > 1\) has only one solution: \(x = 3\), \(a = 2\), \(b = 3\).
#### Proof of Lemma:
1. **Case 1: \(a = 2m\)**
... | (a, b, c) = (3 \cdot 2^{1008}, 2, 2019) | 74,468 | <think>We are given the equation \(2^c + 2^{2016} = a^b\) and need to determine all triples of natural numbers \((a, b, c)\) with \(b > 1\).
### Lemma:
The equation \(2^b + 1 = x^a\) with \(x > 0\) and \(a, b > 1\) has only one solution: \(x = 3\), \(a = 2\), \(b = 3\).
#### Proof of Lemma:
1. **Case 1: \(a = 2m\)**
... | 1_1 |
Let $n \geq 2$ be a positive integer. Each cell of an $n \times n$ board is colored red or blue. We place dominoes on the board, each covering two cells. We call a domino plain if it lies on two red or two blue cells, and colorful if it lies on one red and one blue cell. Find the largest positive integer $k$ with the f... | <think>We prove that $k=\left\lfloor\frac{n^{2}}{4}\right\rfloor$ is the largest possible value.
Suppose $n$ is even. Then it is possible to cover the board with $\frac{n^{2}}{2}$ dominoes (ignoring colors). Since there are $\frac{n^{2}}{2}$ dominoes, all of which are either colorful or plain, there are at least $\left... | \left\lfloor\frac{n^{2}}{4}\right\rfloor | 43,480 | <think>We prove that $k=\left\lfloor\frac{n^{2}}{4}\right\rfloor$ is the largest possible value.
Suppose $n$ is even. Then it is possible to cover the board with $\frac{n^{2}}{2}$ dominoes (ignoring colors). Since there are $\frac{n^{2}}{2}$ dominoes, all of which are either colorful or plain, there are at least $\left... | 1_2 |
Twenty kilograms of cheese are on sale in a grocery store. Several customers are lined up to buy this cheese. After a while, having sold the demanded portion of cheese to the next customer, the salesgirl calculates the average weight of the portions of cheese already sold and declares the number of customers for whom t... | <think>1. Let \( S_n \) be the sum of the weights of cheese bought after the \( n \)-th customer. We need to determine if the salesgirl can declare, after each of the first 10 customers, that there is just enough cheese for the next 10 customers if each customer buys a portion of cheese equal to the average weight of t... | 10 | 50,140 | <think>1. Let \( S_n \) be the sum of the weights of cheese bought after the \( n \)-th customer. We need to determine if the salesgirl can declare, after each of the first 10 customers, that there is just enough cheese for the next 10 customers if each customer buys a portion of cheese equal to the average weight of t... | 1_3 |
Three regular solids: tetrahedron, hexahedron, and octahedron have equal surface areas. What is the ratio of their volumes? | <think>I. solution: Let the edges of the bodies be denoted by $a, b, c$, their surface areas by $F_{1}, F_{2}, F_{3}$, and their volumes by $k_{1}, k_{2}, k_{3}$.
$$
F_{1}=4 \frac{a^{2}}{4} \sqrt{3}=a^{2} \sqrt{3}, \quad F_{2}=6 b^{2}, \quad F_{3}=8 \frac{c^{2}}{4} \sqrt{3}=2 c^{2} \sqrt{3}
$$
According to the proble... | 1:\sqrt[4]{3}:\sqrt[4]{4} | 29,919 | <think>I. solution: Let the edges of the bodies be denoted by $a, b, c$, their surface areas by $F_{1}, F_{2}, F_{3}$, and their volumes by $k_{1}, k_{2}, k_{3}$.
$$
F_{1}=4 \frac{a^{2}}{4} \sqrt{3}=a^{2} \sqrt{3}, \quad F_{2}=6 b^{2}, \quad F_{3}=8 \frac{c^{2}}{4} \sqrt{3}=2 c^{2} \sqrt{3}
$$
According to the proble... | 1_4 |
Example 7 Given a real number $c \in\left(\frac{1}{2}, 1\right)$. Find the smallest constant $M$, such that for any integer $n \geqslant 2$ and real numbers $0<a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$, as long as they satisfy
$$
\frac{1}{n} \sum_{k=1}^{n} k a_{k}=c \sum_{k=1}^{n} a_{k},
$$
we always hav... | <think>Proof: From the given conditions, we have
$$
\sum_{k=1}^{m}(c n-k) a_{k}=\sum_{k=m+1}^{n}(k-c n) a_{k} \text {. }
$$
Since $\frac{1}{2}c n \geqslant m \geqslant 1$.
Notice that $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{m}$,
$$
c n-1 \geqslant c n-2 \geqslant \cdots \geqslant c n-m \text {. }
$$
By C... | \frac{1}{1-c} | 38,162 | <think>Proof: From the given conditions, we have
$$
\sum_{k=1}^{m}(c n-k) a_{k}=\sum_{k=m+1}^{n}(k-c n) a_{k} \text {. }
$$
Since $\frac{1}{2}c n \geqslant m \geqslant 1$.
Notice that $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{m}$,
$$
c n-1 \geqslant c n-2 \geqslant \cdots \geqslant c n-m \text {. }
$$
By C... | 1_5 |
Let's determine the pairs of positive integers $\left(a_{1} ; a_{2}\right)$ for which the sequence defined by the recursion $a_{n+2}=\frac{a_{n}+a_{n+1}}{\left(a_{n}, a_{n+1}\right)}(n \geq 1)$ is periodic. | <think>Solution. The terms of the sequence are clearly positive integers due to the formation rule. Let $(for all n)$ $d_{n}=(a_{n}, a_{n+1})$, then $d_{n+1}$ divides $a_{n+1}$ and $a_{n+2}$; therefore, it also divides $d_{n} a_{n+2}-a_{n+1}=a_{n}$. Thus, since $d_{n+1} \mid a_{n}$ and $d_{n+1} \mid a_{n+1}$, $d_{n+1}$... | a_{1}=a_{2}=2 | 80,406 | <think>Solution. The terms of the sequence are clearly positive integers due to the formation rule. Let $(for all n)$ $d_{n}=(a_{n}, a_{n+1})$, then $d_{n+1}$ divides $a_{n+1}$ and $a_{n+2}$; therefore, it also divides $d_{n} a_{n+2}-a_{n+1}=a_{n}$. Thus, since $d_{n+1} \mid a_{n}$ and $d_{n+1} \mid a_{n+1}$, $d_{n+1}$... | 1_6 |
$\quad$ Inscribed and Circumscribed Circles [ Trigonometric Ratios in a Right Triangle
A regular hexagon $A B C D E K$ is inscribed in a circle of radius $3+2 \sqrt{3}$. Find the radius of the circle inscribed in triangle $B C D$.
# | <think>The radius of the circle inscribed in a triangle is equal to the area of the triangle divided by its semiperimeter.
## Solution
Let $R$ be the radius of the given circle. In triangle $BCD$, it is known that
$$
\begin{gathered}
\angle BCD = 120^\circ, BC = CD = R, BD = R \sqrt{3} \\
S_{\triangle BCD} = \frac{1... | \frac{3}{2} | 70,467 | <think>The radius of the circle inscribed in a triangle is equal to the area of the triangle divided by its semiperimeter.
## Solution
Let $R$ be the radius of the given circle. In triangle $BCD$, it is known that
$$
\begin{gathered}
\angle BCD = 120^\circ, BC = CD = R, BD = R \sqrt{3} \\
S_{\triangle BCD} = \frac{1... | 1_7 |
Let $O$ and $I$ be the circumcenter and incenter of triangle $ABC$. The perpendicular from $I$ to $OI$ meets $AB$ and the external bisector of angle $C$ at points $X$ and $Y$ respectively. In what ratio does $I$ divide the segment $XY$? | <think>1. **Define the problem and setup:**
- Let \( O \) be the circumcenter and \( I \) be the incenter of triangle \( ABC \).
- The perpendicular from \( I \) to \( OI \) meets \( AB \) at \( X \) and the external bisector of angle \( C \) at \( Y \).
- We need to find the ratio in which \( I \) divides the... | 1:2 | 87,546 | <think>1. **Define the problem and setup:**
- Let \( O \) be the circumcenter and \( I \) be the incenter of triangle \( ABC \).
- The perpendicular from \( I \) to \( OI \) meets \( AB \) at \( X \) and the external bisector of angle \( C \) at \( Y \).
- We need to find the ratio in which \( I \) divides the... | 1_8 |
Let $x_1,x_2,x_3,\dots$ be sequence of nonzero real numbers satisfying $$x_n=\frac{x_{n-2}x_{n-1}}{2x_{n-2}-x_{n-1}}, \quad \quad n=3,4,5,\dots$$ Establish necessary and sufficient conditions on $x_1,x_2$ for $x_n$ to be an integer for infinitely many values of $n$. | <think>1. We start with the given recurrence relation for the sequence \( \{x_n\} \):
\[
x_n = \frac{x_{n-2} x_{n-1}}{2x_{n-2} - x_{n-1}}, \quad n = 3, 4, 5, \dots
\]
2. We aim to find a general form for \( x_n \). Let's assume \( x_1 \) and \( x_2 \) are given. We will use induction to find a pattern.
3. Fo... | x_1 = x_2 \in \mathbb{Z} \setminus \{0\} | 25,477 | <think>1. We start with the given recurrence relation for the sequence \( \{x_n\} \):
\[
x_n = \frac{x_{n-2} x_{n-1}}{2x_{n-2} - x_{n-1}}, \quad n = 3, 4, 5, \dots
\]
2. We aim to find a general form for \( x_n \). Let's assume \( x_1 \) and \( x_2 \) are given. We will use induction to find a pattern.
3. Fo... | 1_9 |
Let $n>1$ be a natural number. Find the real values of the parameter $a$, for which the equation $\sqrt[n]{1+x}+\sqrt[n]{1-x}=a$ has a single real root. | <think>1. Consider the given equation:
\[
\sqrt[n]{1+x} + \sqrt[n]{1-x} = a
\]
where \( n > 1 \) is a natural number and \( a \) is a real parameter.
2. Define the function:
\[
f(x) = \sqrt[n]{1+x} + \sqrt[n]{1-x}
\]
We need to analyze the behavior of \( f(x) \) to determine for which values of... | a = 2 | 31,521 | <think>1. Consider the given equation:
\[
\sqrt[n]{1+x} + \sqrt[n]{1-x} = a
\]
where \( n > 1 \) is a natural number and \( a \) is a real parameter.
2. Define the function:
\[
f(x) = \sqrt[n]{1+x} + \sqrt[n]{1-x}
\]
We need to analyze the behavior of \( f(x) \) to determine for which values of... | 1_10 |
Find $3x^2 y^2$ if $x$ and $y$ are integers such that $y^2 + 3x^2 y^2 = 30x^2 + 517$. | <think>1. Given the equation \( y^2 + 3x^2 y^2 = 30x^2 + 517 \), we start by letting \( a = x^2 \) and \( b = y^2 \). This transforms the equation into:
\[
b + 3ab = 30a + 517
\]
2. We can use Simon's Favorite Factoring Trick to factor the equation. First, we rearrange the terms:
\[
b + 3ab - 30a = 517
... | 588 | 44,602 | <think>1. Given the equation \( y^2 + 3x^2 y^2 = 30x^2 + 517 \), we start by letting \( a = x^2 \) and \( b = y^2 \). This transforms the equation into:
\[
b + 3ab = 30a + 517
\]
2. We can use Simon's Favorite Factoring Trick to factor the equation. First, we rearrange the terms:
\[
b + 3ab - 30a = 517
... | 1_11 |
The sequence $(x_n)_{n\geqslant 0}$ is defined as such: $x_0=1, x_1=2$ and $x_{n+1}=4x_n-x_{n-1}$, for all $n\geqslant 1$. Determine all the terms of the sequence which are perfect squares.
[i]George Stoica, Canada[/i] | <think>1. **Define the sequence and initial conditions:**
The sequence \((x_n)_{n \geq 0}\) is defined by:
\[
x_0 = 1, \quad x_1 = 2, \quad \text{and} \quad x_{n+1} = 4x_n - x_{n-1} \quad \text{for all} \quad n \geq 1.
\]
2. **Identify the recurrence relation:**
The recurrence relation is:
\[
x_{n... | x_0 = 1 | 8,869 | <think>1. **Define the sequence and initial conditions:**
The sequence \((x_n)_{n \geq 0}\) is defined by:
\[
x_0 = 1, \quad x_1 = 2, \quad \text{and} \quad x_{n+1} = 4x_n - x_{n-1} \quad \text{for all} \quad n \geq 1.
\]
2. **Identify the recurrence relation:**
The recurrence relation is:
\[
x_{n... | 1_12 |
Find all natural numbers $n, k$ such that
$$ 2^n – 5^k = 7. $$ | <think>To find all natural numbers \( n \) and \( k \) such that \( 2^n - 5^k = 7 \), we can proceed as follows:
1. **Initial Check for Small Values of \( n \):**
- For \( n = 1 \):
\[
2^1 - 5^k = 2 - 5^k = 7 \implies 5^k = -5 \quad \text{(no solution since \( 5^k \) is positive)}
\]
- For \( n = ... | (n, k) = (5, 2) | 25,628 | <think>To find all natural numbers \( n \) and \( k \) such that \( 2^n - 5^k = 7 \), we can proceed as follows:
1. **Initial Check for Small Values of \( n \):**
- For \( n = 1 \):
\[
2^1 - 5^k = 2 - 5^k = 7 \implies 5^k = -5 \quad \text{(no solution since \( 5^k \) is positive)}
\]
- For \( n = ... | 1_13 |
B1. At least how many times must we roll two fair dice simultaneously so that the probability of rolling the same number of pips on both dice at least once is greater than $\frac{1}{2}$?
(20 points) | <think>B1. The power of the algebra of events when rolling two dice is $6 \cdot 6=36$. At the same time, the same number of dots can fall on both dice in 6 ways. The probability that the same number of dots falls on both dice when rolled once is therefore $p=\frac{6}{36}=\frac{1}{6}$.
Suppose we roll the dice $n$ time... | 4 | 48,727 | <think>B1. The power of the algebra of events when rolling two dice is $6 \cdot 6=36$. At the same time, the same number of dots can fall on both dice in 6 ways. The probability that the same number of dots falls on both dice when rolled once is therefore $p=\frac{6}{36}=\frac{1}{6}$.
Suppose we roll the dice $n$ time... | 1_14 |
16. Find all positive integers $n$ such that the equation
$$
\frac{1}{x_{1}^{2}}+\frac{1}{x_{2}^{2}}+\cdots+\frac{1}{x_{n}^{2}}=\frac{n+1}{x_{n+1}^{2}}
$$
has positive integer solutions. | <think>16. (1) When $n=1$, the equation becomes
$$
\frac{1}{x_{1}^{2}}=\frac{2}{x_{2}^{2}} \Rightarrow \frac{x_{2}}{x_{1}}=\sqrt{2} \text {, }
$$
obviously there is no positive integer solution.
(2) When $n=2$, the equation becomes
$$
\begin{array}{l}
\frac{1}{x_{1}^{2}}+\frac{1}{x_{2}^{2}}=\frac{3}{x_{3}^{2}} \\
\Rig... | n\geqslant3 | 71,394 | <think>16. (1) When $n=1$, the equation becomes
$$
\frac{1}{x_{1}^{2}}=\frac{2}{x_{2}^{2}} \Rightarrow \frac{x_{2}}{x_{1}}=\sqrt{2} \text {, }
$$
obviously there is no positive integer solution.
(2) When $n=2$, the equation becomes
$$
\begin{array}{l}
\frac{1}{x_{1}^{2}}+\frac{1}{x_{2}^{2}}=\frac{3}{x_{3}^{2}} \\
\Rig... | 1_15 |
Example 3. Solve the equation
$$
2 f(x+y)+f(x-y)=f(x)\left(2 e^{y}+e^{-y}\right)
$$
28 | <think>Solution. Performing the following substitutions:
$$
x=0, y=t ; x=t, y=2 t ; x=t, y=-2 t
$$
we obtain the equations:
$$
\begin{aligned}
& 2 f(t)+f(-t)=a\left(2 e^{t}+e^{-t}\right) \\
& 2 f(3 t)+f(-t)=f(t)\left(2 e^{2 t}+e^{-2 t}\right) \\
& 2 f(-t)+f(3 t)=f(t)\left(2 e^{-2 t}+e^{2 t}\right)
\end{aligned}
$$
... | f()=\cdote^{} | 3,469 | <think>Solution. Performing the following substitutions:
$$
x=0, y=t ; x=t, y=2 t ; x=t, y=-2 t
$$
we obtain the equations:
$$
\begin{aligned}
& 2 f(t)+f(-t)=a\left(2 e^{t}+e^{-t}\right) \\
& 2 f(3 t)+f(-t)=f(t)\left(2 e^{2 t}+e^{-2 t}\right) \\
& 2 f(-t)+f(3 t)=f(t)\left(2 e^{-2 t}+e^{2 t}\right)
\end{aligned}
$$
... | 1_16 |
There are 15 cities, and there is a train line between each pair operated by either the Carnegie Rail Corporation or the Mellon Transportation Company. A tourist wants to visit exactly three cities by travelling in a loop, all by travelling on one line. What is the minimum number of such 3-city loops? | <think>1. **Define the problem in terms of graph theory**:
- We have 15 cities, which we can represent as vertices in a complete graph \( K_{15} \).
- Each edge in this graph is colored either by the Carnegie Rail Corporation (CRC) or the Mellon Transportation Company (MTC).
- We need to find the minimum numbe... | 88 | 29,762 | <think>1. **Define the problem in terms of graph theory**:
- We have 15 cities, which we can represent as vertices in a complete graph \( K_{15} \).
- Each edge in this graph is colored either by the Carnegie Rail Corporation (CRC) or the Mellon Transportation Company (MTC).
- We need to find the minimum numbe... | 1_17 |
In the game of [i]Winners Make Zeros[/i], a pair of positive integers $(m,n)$ is written on a sheet of paper. Then the game begins, as the players make the following legal moves:
[list]
[*] If $m\geq n$, the player choose a positive integer $c$ such that $m-cn\geq 0$, and replaces $(m,n)$ with $(m-cn,n)$.
[*] If $m<n$... | <think>To solve this problem, we need to analyze the game and determine the largest choice the first player can make for \( c \) such that the first player has a winning strategy after that first move. We start with the initial pair \((m, n) = (2007777, 2007)\).
1. **Initial Setup and First Move**:
- Given \( m = 2... | 999 | 15,493 | <think>To solve this problem, we need to analyze the game and determine the largest choice the first player can make for \( c \) such that the first player has a winning strategy after that first move. We start with the initial pair \((m, n) = (2007777, 2007)\).
1. **Initial Setup and First Move**:
- Given \( m = 2... | 1_18 |
Find all functions $f:\mathbb{N}\to\mathbb{N}$ such that for all $x,y\in\mathbb{N}$:
$$0\le y+f(x)-f^{f(y)}(x)\le1$$
that here
$$f^n(x)=\underbrace{f(f(\ldots(f}_{n}(x))\ldots)$$ | <think>1. **Lemma 1.0**: For each natural number \( x \), the set \(\mathcal{O}(x) = \{f^n(x) : n \ge 0\}\) is unbounded.
- Proof: For each natural number \( y \), we have:
\[
f^{f(y)}(x) \ge y + f(x) - 1
\]
This implies that as \( y \) increases, \( f^{f(y)}(x) \) can take arbitrarily large valu... | f(n) = n + 1 | 65,403 | <think>1. **Lemma 1.0**: For each natural number \( x \), the set \(\mathcal{O}(x) = \{f^n(x) : n \ge 0\}\) is unbounded.
- Proof: For each natural number \( y \), we have:
\[
f^{f(y)}(x) \ge y + f(x) - 1
\]
This implies that as \( y \) increases, \( f^{f(y)}(x) \) can take arbitrarily large valu... | 1_19 |
Question 53: Find all prime pairs $(p, q)$ such that $\left(3 p^{q-1}+1\right) \mid\left(11^{p}+17^{p}\right)$. | <think>Question 53:
Solution: If $p=2$, then $\left(3 \cdot 2^{q-1}+1\right) \mid\left(11^{2}+17^{2}\right)$, which means $\left(3 \cdot 2^{q-1}+1\right) \mid 2 \times 5 \times 41$. Upon inspection, there is no solution.
Now assume $p \geq 3$, which is an odd prime number.
Notice that $11^{p}+17^{p} \equiv 11+17 \equiv... | (p,q)=(3,3) | 19,368 | <think>Question 53:
Solution: If $p=2$, then $\left(3 \cdot 2^{q-1}+1\right) \mid\left(11^{2}+17^{2}\right)$, which means $\left(3 \cdot 2^{q-1}+1\right) \mid 2 \times 5 \times 41$. Upon inspection, there is no solution.
Now assume $p \geq 3$, which is an odd prime number.
Notice that $11^{p}+17^{p} \equiv 11+17 \equiv... | 1_20 |
Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$, respectively, and are externally tangent at point $A$. Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and in... | <think>Let $M$ be the intersection of $\overline{BC}$ and the common internal tangent of $\mathcal P$ and $\mathcal Q.$ We claim that $M$ is the circumcenter of right $\triangle{ABC}.$ Indeed, we have $AM = BM$ and $BM = CM$ by equal tangents to circles, and since $BM = CM, M$ is the midpoint of $\overline{BC},$ implyi... | 129 | 16,599 | <think>Let $M$ be the intersection of $\overline{BC}$ and the common internal tangent of $\mathcal P$ and $\mathcal Q.$ We claim that $M$ is the circumcenter of right $\triangle{ABC}.$ Indeed, we have $AM = BM$ and $BM = CM$ by equal tangents to circles, and since $BM = CM, M$ is the midpoint of $\overline{BC},$ implyi... | 1_21 |
Exercise 12. Let $(a_n)$ be a sequence of real numbers. Suppose that $a_0 = 1$ and for all $n \geqslant 1, a_n$ is the smallest strictly positive solution of
$$
\left(a_n - a_{n-1}\right)\left(a_n + a_{n-1} - 2 \sqrt{n}\right) = 2
$$
Find the smallest integer $n$ such that $a_n \geqslant 2022$. | <think>Solution to Exercise 12 To get an idea of the problem, we calculate the first values of $a_{n}$: for $a_{1}$, the equation is
$$
\left(a_{1}-1\right)\left(a_{1}-1\right)=2
$$
so $a_{1}-1=\sqrt{2}$ or $-\sqrt{2}$. Only the first case gives a positive solution $a_{1}=1+\sqrt{2}$. Now let's try to calculate $a_{2... | 1011^2 | 61,599 | <think>Solution to Exercise 12 To get an idea of the problem, we calculate the first values of $a_{n}$: for $a_{1}$, the equation is
$$
\left(a_{1}-1\right)\left(a_{1}-1\right)=2
$$
so $a_{1}-1=\sqrt{2}$ or $-\sqrt{2}$. Only the first case gives a positive solution $a_{1}=1+\sqrt{2}$. Now let's try to calculate $a_{2... | 1_22 |
Let $ z = \frac{1}{2}(\sqrt{2} + i\sqrt{2}) $. The sum $$ \sum_{k = 0}^{13} \dfrac{1}{1 - ze^{k \cdot \frac{i\pi}{7}}} $$
can be written in the form $ a - bi $. Find $ a + b $. | <think>1. Given \( z = \frac{1}{2}(\sqrt{2} + i\sqrt{2}) \), we first simplify \( z \):
\[
z = \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}
\]
This can be written as:
\[
z = \frac{1}{\sqrt{2}} (1 + i)
\]
2. We need to find the sum:
\[
\sum_{k=0}^{13} ... | 7 - 7i | 40,026 | <think>1. Given \( z = \frac{1}{2}(\sqrt{2} + i\sqrt{2}) \), we first simplify \( z \):
\[
z = \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}
\]
This can be written as:
\[
z = \frac{1}{\sqrt{2}} (1 + i)
\]
2. We need to find the sum:
\[
\sum_{k=0}^{13} ... | 1_23 |
9. 56 Find
$$\begin{aligned}
A & =\sqrt{\left(1264-z_{1}-\cdots-z_{n}\right)^{2}+x_{n}^{2}+y_{n}^{2}}+ \\
& \sqrt{z_{n}^{2}+x_{n-1}^{2}+y_{n-1}^{2}}+\cdots+\sqrt{z_{2}^{2}+x_{1}^{2}+y_{1}^{2}}+ \\
& \sqrt{z_{1}^{2}+\left(948-x_{1}-\cdots-x_{n}\right)^{2}+\left(1185-y_{1}-\cdots-y_{n}\right)^{2}}
\end{aligned}$$
the mi... | <think>[Solution] Consider the following points in a three-dimensional space with a Cartesian coordinate system:
$$\begin{array}{l}
B=(0,0,1264) \\
M_{n}=\left(x_{n}, y_{n}, z_{1}+\cdots+z_{n}\right) \\
M_{n-1}=\left(x_{n-1}+x_{n}, y_{n-1}+y_{n}, z_{1}+\cdots+z_{n-1}\right) \\
\cdots \cdots \cdots \cdots \\
M_{2}=\left... | 1975 | 6,145 | <think>[Solution] Consider the following points in a three-dimensional space with a Cartesian coordinate system:
$$\begin{array}{l}
B=(0,0,1264) \\
M_{n}=\left(x_{n}, y_{n}, z_{1}+\cdots+z_{n}\right) \\
M_{n-1}=\left(x_{n-1}+x_{n}, y_{n-1}+y_{n}, z_{1}+\cdots+z_{n-1}\right) \\
\cdots \cdots \cdots \cdots \\
M_{2}=\left... | 1_24 |
In a culturing of bacteria, there are two species of them: red and blue bacteria.
When two red bacteria meet, they transform into one blue bacterium.
When two blue bacteria meet, they transform into four red bacteria.
When a red and a blue bacteria meet, they transform into three red bacteria.
Find, in function of th... | <think>To solve this problem, we need to analyze the transitions between different states of the bacteria population. We will use the given transformation rules and the concept of Markov chains to determine the possible amounts of bacteria and the possible amounts of red and blue bacteria for each possible total amount... | \{(n, 0), (n-2, 1), (n-4, 2), \ldots\} | 76,474 | <think>To solve this problem, we need to analyze the transitions between different states of the bacteria population. We will use the given transformation rules and the concept of Markov chains to determine the possible amounts of bacteria and the possible amounts of red and blue bacteria for each possible total amount... | 1_25 |
## Task A-2.4.
Two circles that do not intersect are given, with radii $r_{1}$ and $r_{2}$. The distance between the points of tangency of the common internal tangent to these circles is 12, and the distance between the points of tangency of the common external tangent to these circles is 16. Determine the product $r_... | <think>## First Solution.
Let points $A$ and $B$ be the points of tangency of the external tangent, and points $P$ and $Q$ be the points of tangency of the internal tangent with the given circles of radii $r_{1}$ and $r_{2}$, respectively. Let $S_{1}$ be the center of the circle with radius $r_{1}$ and $S_{2}$ be the ... | 28 | 5,374 | <think>## First Solution.
Let points $A$ and $B$ be the points of tangency of the external tangent, and points $P$ and $Q$ be the points of tangency of the internal tangent with the given circles of radii $r_{1}$ and $r_{2}$, respectively. Let $S_{1}$ be the center of the circle with radius $r_{1}$ and $S_{2}$ be the ... | 1_26 |
A set $A$ is endowed with a binary operation $*$ satisfying the following four conditions:
(1) If $a, b, c$ are elements of $A$, then $a * (b * c) = (a * b) * c$ ,
(2) If $a, b, c$ are elements of $A$ such that $a * c = b *c$, then $a = b$ ,
(3) There exists an element $e$ of $A$ such that $a * e = a$ for all $a$ in $A... | <think>To determine the largest cardinality that the set \( A \) can have, we need to analyze the given conditions carefully.
1. **Associativity**: The operation \( * \) is associative, i.e., for all \( a, b, c \in A \),
\[
a * (b * c) = (a * b) * c.
\]
2. **Cancellation Law**: If \( a * c = b * c \), then \... | 3 | 79,460 | <think>To determine the largest cardinality that the set \( A \) can have, we need to analyze the given conditions carefully.
1. **Associativity**: The operation \( * \) is associative, i.e., for all \( a, b, c \in A \),
\[
a * (b * c) = (a * b) * c.
\]
2. **Cancellation Law**: If \( a * c = b * c \), then \... | 1_27 |
4. Krt Črt has 5 rooms in his burrow, numbered from 1 to 5. Between some of them, Črt has dug tunnels, so from each room he can slide through several tunnels to any other room. No two tunnels intersect. Each tunnel starts in one room, ends in another (different from the starting one), and does not pass through any room... | <think>III/4. Room 5 is not adjacent to room 1, since Črt could otherwise walk from room 5 to room 1 three times, first from room 5 to room 1, then back to room 5, and then again to room 1. This would mean he would reach room 1 from room 5 in exactly three tunnels, which contradicts the problem statement. Room 5 is als... | (1,4),(2,3),(2,4),(3,4),(4,5) | 79,172 | <think>III/4. Room 5 is not adjacent to room 1, since Črt could otherwise walk from room 5 to room 1 three times, first from room 5 to room 1, then back to room 5, and then again to room 1. This would mean he would reach room 1 from room 5 in exactly three tunnels, which contradicts the problem statement. Room 5 is als... | 1_28 |
Two containers, a bucket and a basin, have volumes of 3 and 5 liters, respectively. By removing water from a lake, how can we leave the basin with exactly 4 liters of water using only these two containers?
$, where $a$ indicates the number of liters in the bucket and $b$ in the basin. At the beginning, when both containers are empty, we have the pair $(0,0)$. At the stage, we want to obtain the pair $(a, 4)$, wh... | (3,4) | 63,029 | <think>Solution
At any time, we can indicate the amounts of water in the two containers through a pair $(a, b)$, where $a$ indicates the number of liters in the bucket and $b$ in the basin. At the beginning, when both containers are empty, we have the pair $(0,0)$. At the stage, we want to obtain the pair $(a, 4)$, wh... | 1_29 |
On each OMA lottery ticket there is a $9$-digit number that only uses the digits $1, 2$ and $3$ (not necessarily all three). Each ticket has one of the three colors red, blue or green. It is known that if two banknotes do not match in any of the $9$ figures, then they are of different colors. Bill $122222222$ is red, $... | <think>1. **Identify the given information:**
- Ticket $122222222$ is red.
- Ticket $222222222$ is green.
- If two tickets do not match in any of the 9 digits, then they are of different colors.
2. **Determine the color of ticket $311311311$:**
- Compare $311311311$ with $122222222$:
- $311311311$ and... | \text{red} | 19,166 | <think>1. **Identify the given information:**
- Ticket $122222222$ is red.
- Ticket $222222222$ is green.
- If two tickets do not match in any of the 9 digits, then they are of different colors.
2. **Determine the color of ticket $311311311$:**
- Compare $311311311$ with $122222222$:
- $311311311$ and... | 1_30 |
Mad scientist Kyouma writes $N$ positive integers on a board. Each second, he chooses two numbers $x, y$ written on the board with $x > y$, and writes the number $x^2-y^2$ on the board. After some time, he sends the list of all the numbers on the board to Christina. She notices that all the numbers from 1 to 1000 are p... | <think>1. **Define the Set \( S \):**
We start by defining the set \( S \) of numbers initially written on the board. Let
\[
X = \{x: x = 4k+2, k \in \mathbb{Z}, 0 \le k \le 249\}.
\]
Then,
\[
S = \{1, 4\} \cup X.
\]
This set \( S \) contains 252 elements: 250 elements from \( X \) and the ... | N = 252 | 50,335 | <think>1. **Define the Set \( S \):**
We start by defining the set \( S \) of numbers initially written on the board. Let
\[
X = \{x: x = 4k+2, k \in \mathbb{Z}, 0 \le k \le 249\}.
\]
Then,
\[
S = \{1, 4\} \cup X.
\]
This set \( S \) contains 252 elements: 250 elements from \( X \) and the ... | 1_31 |
Equilateral $\triangle ABC$ has side length $\sqrt{111}$. There are four distinct triangles $AD_1E_1$, $AD_1E_2$, $AD_2E_3$, and $AD_2E_4$, each congruent to $\triangle ABC$,
with $BD_1 = BD_2 = \sqrt{11}$. Find $\sum_{k=1}^4(CE_k)^2$. | <think>Note that there are only two possible locations for points $D_1$ and $D_2$, as they are both $\sqrt{111}$ from point $A$ and $\sqrt{11}$ from point $B$, so they are the two points where a circle centered at $A$ with radius $\sqrt{111}$ and a circle centered at $B$ with radius $\sqrt{11}$ intersect. Let $D_1$ be... | 677 | 52,783 | <think>Note that there are only two possible locations for points $D_1$ and $D_2$, as they are both $\sqrt{111}$ from point $A$ and $\sqrt{11}$ from point $B$, so they are the two points where a circle centered at $A$ with radius $\sqrt{111}$ and a circle centered at $B$ with radius $\sqrt{11}$ intersect. Let $D_1$ be... | 1_32 |
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